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Internal Direct Product

“Sometimes, I don’t understand what you’re talking about,” I said, shrugging my shoulders. “Most of the time, I don’t understand what I’m talking about. There is nothing more unknown as a sure thing. The future is no more uncertain than the present. The present is a chaotic masquerade; the world, a meaningless merry-go-round, life is a messy game, and I am just a broken, pathetic pawn. I believe that I know nothing, and yet I am not quite sure about that,” Apocalypse, Anawim, #justtothepoint.

Recall. Let G1, G2, . . . , Gn be a finite collection of groups. The external direct product of G1, G2, . . . , Gn, written as G1 ⊕ G2 ⊕ ··· ⊕ Gn, is the set of all n-tuples for which the ith component is an element of Gi and the operation is component-wise, e.g., ℤ3 ⊕ U(5) ⊕ D4, (2, 3, sr)(2, 4, sr2) = (2+2, 3·4, (sr)(sr2)) =[3·4 = 12 (mod 5) = 2; (sr)(sr2) = s(rs)r2 = s(sr-1)r2 = s(sr3)r2 = s2r5 = r] (1, 2, r); ℤ2 ⊕ ℤ2 ⊕ ℤ2, (1, 0, 1) + (0, 1, 0) = (1, 1, 1).

Let G be a group, H and H subgroups of G (H, K ≤ G), G is the internal direct product of H and K, G = H x K or G = HK (the notation is unfortunately not standard) if

  1. H and K are normal subgroups of G, H, K ◁ G.
  2. Their intersection is trivial, H ∩ K = {e}.
  3. Every element of G can be expressed uniquely as the product of an element of an element of H and K, i.e, G = HK = {hk | h ∈ H, k ∈ K}.
    Most authors, they only require that G is the subset product of H and K, that is, G = HK.

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Examples

Let’s check that H ◁ G ↭ ∀g ∈ D6, g·H = H·g. Of course, g·e = e·g = g. Besides, g = rk, r3 g = r3rk = r3+k = rk+3 = rkr3 = gr3. If g = rks, r3g = r3rks = rk(r3s) =[Dihedral group, rks = srn-k, 1≤k≤n-1. In particular, k = 3, r3s = sr6-3 = sr3] rk(sr3) = (rks)r3 = gr3

Recall that Z(G) = {z ∈ G| zg = gz ∀g ∈G}. Z(D6) = ⟨r3⟩ = {e, r3}.

More generally, let H1, H2, ···, Hn be a finite collection of subgroups of G. G is the internal direct product of H1, H2, ··· Hn, G = H1 x H2 x ··· x Hn, if

  1. Hi ◁ G, ∀i, 1 ≤ i ≤ n.
  2. Every element of G can be expressed as h1h2···hn. G = H1 H2 ··· Hn = {h1h2···hn | hi ∈ Hi}
  3. Each subgroup is disjoint (with the exception of the identity) from the product of all previous subgroups, (H1H2···Hi) ∩ Hi+1 = {e} ∀i, 1 ≤ i ≤ n-1.

Theorem. If G is the internal direct product of a finite number of subgroups H1, H2, ··· Hn, then G is isomorphic to the external direct product of H1, H2, ··· Hn. More concisely, G = H1 x H2 x ··· x Hn ≋ H1 ⊕ H2 ⊕ ··· ⊕ Hn

Proof.

Let G = H1 x H2 x ··· x Hn

[By assumption, Hj ◁ G ↭ gHjg-1 ⊆ Hj, let g =G=H1xH2x···xHn (e·e···hi···e·e) = hi] hihjhi-1 ∈ Hj and (hihjhi-1)hj-1 ∈ Hjhj-1 ⇒ [Ha = H ↭ a ∈ H, hj-1 ∈ Hj] hihjhi-1hj-1 ∈ Hjhj-1 = Hj and similarly hi(hjhi-1hj-1) ∈ hiHi = Hi.

Therefore, hihjhi-1hj-1 ∈ Hi ∩ Hj.

i ≠ j, say i < j, let h ∈ Hi ∩ Hj, h = e1···ei-1hei+1···ej-1, therefore h ∈ H1 ··· Hi ··· Hj-1 ∩ Hj =[Condition 3. Each subgroup is disjoint from the product of all previous subgroup.] {e}.

hihjhi-1hj-1 ∈ Hi ∩ Hj = {e} ⇒ hihjhi-1hj-1 = e ⇒ hihj = hjhi

Suppose, h1h2···hn = h’1h’2···h’n ⇒ h1h2···hnhn-1 = h’1h’2···h’nhn-1 ⇒ h1h2···hn-1 = h’1h’2···h’nhn-1 ⇒ h1h2···hn-1hn-1-1 = h’1h’2···h’nhn-1hn-1-1 ⇒ [∀hi ∈ Hi, hj ∈ Hj, i ≠ j, hihj = hjhi] h1h2···hn-2 = h’1h’2···hn-1h’n-1-1h’nhn-1 ⇒ e = h’1h1-1h’2h2-1 ··· h’nhn-1 ⇒ hn = h’1h1-1h’2h2-1 ··· h’n ⇒ [Multiplying both sides by h’n-1 and ∀hi ∈ Hi, hj ∈ Hj, i ≠ j, hihj = hjhi] h’n-1hn = h’1h1-1 ··· h’n-1hn-1-1 ∈ H1H2 ··· Hn-1 ∩ Hn = {e} ⇒ h’n-1hn = e ⇒ h’n = hn. Therefore, h1h2···hn-1 = h’1h’2···h’n-1 and repeating the argument, we can obtain hn-1 = h’n-1, and so on, so forth hi = h’i ∀i, 1 ≤ i ≤ n.

  1. Since every element of G has an unique representation as h1h2···hn, Φ is well defined, onto, and one-to-one.
  2. Let a = h1h2···hn, b = h’1h’2···h’n, Φ(ab) = Φ(h1h2···hnh’1h’2···h’n) = [∀hi ∈ Hi, hj ∈ Hj, i ≠ j, hihj = hjhi] Φ(h1h’1h2h’2···hnh’n) = (h1h’1, h2h’2,···, hnh’n) = (h1, h2,···, hn)(h’1, h’2,···, h’n) = Φ(a)·Φ(b)∎

Examples

Recall that ℤ(Dn) = $ \begin{cases} e, nis odd\\\\ e~ and~ α^{\frac{n}{2}}, nis even \end{cases}$

Let H2 = ⟨rn⟩ = {e, rn} = ℤ(D2n) ⇒ [The center of a group is a normal subgroup] H2 ◁ D2n. Futhermore, D2n = H1H2, and H1 ∩ H2 = {e} ⇒ [By the previous theorem] D2n ≋ H1 ⊕ H2 ≋ ℤ2 ⊕ Dn, e.g., D6 ≋ ℤ2 ⊕ D3, D14 ≋ ℤ2 ⊕ D7.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
  8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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